Kerala KEAM 5 June 2024 Solved Paper

We are asked to simplify the expression: $$\frac{1 + (a - ib)}{1 + (a + ib)}.$$Step 1: Simplify the numerator and denominator: $$Numerator: 1 + (a - ib) = 1 + a - ib.$$ $$Denominator: 1 + (a + ib) = 1 + a + ib.$$Thus, the expression becomes: $$\frac{1 + a - ib}{1 + a + ib}.$$Step 2: Multiply both the numerator and denominator by the conjugate of the denominator: $$\frac{(1 + a - ib)}{(1 + a + ib)} \times \frac{(1 + a - ib)}{(1 + a - ib)} = \frac{(1 + a - ib)^2}{(1 + a)^2 + b^2}.$$Step 3: Use the given condition $a^2 + b^2 = 1$. Thus, the denominator simplifies to: $$(1 + a)^2 + b^2 = 1 + 2a + a^2 + b^2 = 1 + 2a + 1 = 2 + 2a.$$Step 4: Now, expand the numerator: $$(1 + a - ib)^2 = (1 + a)^2 - 2i b(1 + a) + (-ib)^2 = (1 + 2a + a^2) - 2i b(1 + a) - b^2.$$Since $b^2 = 1 - a^2$, we get: $$(1 + a - ib)^2 = 1 + 2a + a^2 - 2ib(1 + a) - (1 - a^2) = 2a + 2a^2 - 2ib(1 + a).$$Step 5: Substitute back into the expression: $$\frac{2a + 2a^2 - 2ib(1 + a)}{2 + 2a}.$$Now, factor out the common terms: $$\frac{2a(1 + a) - 2ib(1 + a)}{2(1 + a)} = \frac{2a - 2ib}{2} = a - ib.$$Thus, the correct answer is option (A), $a - ib$. Quick Tip: When simplifying complex rational expressions, multiply both the numerator and denominator by the conjugate of the denominator to eliminate the imaginary part in the denominator.